Equation: $$y = \frac{x + 1}{x - 1}$$
Task: Solve for $x$ in terms of $y$.
My attempt: I just dont know how to go about doing this because the variable $x$ is in both the numerator and denominator.
$\endgroup$ 14 Answers
$\begingroup$You can multiply by $x-1$ on both sides to get $yx-y=x+1$. Then you can move everything to one side $yx-x-y-1=0$. Can you get it from there?
$\endgroup$ 1 $\begingroup$\begin{align} y &= \frac{x+1}{x-1} \\ y(x-1) &= x+1 \\ (y-1) x &= y+1 \\ x &= \frac{y+1}{y-1} \end{align}
$\endgroup$ $\begingroup$Multiply $(x-1)$ on each side to get $y(x-1) = x+1$ which is equal to $yx -y = x + 1$ now subtract $1$ from each side to get $yx -y -1 = x$, now subtract $yx$ from each side to get $-y -1 = x -yx$ which is equal to $-y -1 = x(1 -y)$ Finally , divide by $1-y$ each side to get $$x=\frac{-y -1}{1 - y}$$ and you are done !
$\endgroup$ $\begingroup$we have $$\frac y 1=\frac{x+1}{x-1} \implies \frac{y+1}{y-1} = \frac{(x+1)+(x-1)}{(x+1)-(x-1)}=\frac {2x} 2 = \frac x 1.$$ that is $$x = \frac{y+1}{y-1}. $$
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